An experiment similar to the one in this module was performed to determine the s

ID: 510597 • Letter: A

Question

An experiment similar to the one in this module was performed to determine the solubility and the solubility product constant of gallic acid, a monoprotic organic acid (gmm = 170.12). The equilibrium involved is:

C6H5O3COOH(s) <-----> C6H5O3COO- (aq) + H+(aq)

Titrations using 1.14x10-1 M NaOH were performed, and the following data were obtained

Calculate the following:

What is the average Ksp?

What is the solubility of acid in g per 100 mL?

Chemistry Handbook value for solubility at 20 degrees Celsius: 1.15 g per mL

Determination 1 Determination 2 Determination 3 Temperature of Solution, C 20.2 20.4 20.1 Volume of Acid Solution Titrated, mL 25.00 22.00 20.10 Volume of NaOH solution used, mL 14.61 13.02 11.73

Explanation / Answer

Determination 1

Determination 2

Determination 3

Number of moles of NaOH used = (volume of NaOH used in L)*(concentration of NaOH in mol/L)

(0.01461 L)*(1.14*10-1 mol/L) = 0.001665

(0.01302 L)* (1.14*10-1 mol/L) = 0.001484

(0.01173 L)* (1.14*10-1 mol/L) = 0.001337

Number of moles of H+ titrated = number of moles of NaOH used

0.001665

0.001484

0.001337

[H+] in acid solution = (number of moles of H+ titrated)/(volume of acid taken in L)

(0.001665 mole)/(0.025 L) = 0.0666 M

(0.001484 mole)/(0.022 L) = 0.0674 M

(0.001337 mole)/(0.0201 L) = 0.0665 M

[C6H5O3COO-] in acid solution = [H+] in acid solution (due to the 1:1 nature of dissociation of the acid)

0.0666

0.0674

0.0665

Ksp = [H+][C6H5O3COO-]

(0.0666)*(0.0666) = 4.435*10-3

(0.0674)*(0.0674) = 4.543*10-3

(0.0665)*(0.0665) = 4.422*10-3

Solubility of the acid = [H+] in acid solution = 0.0666 mol/L.

Molar mass of acid = 170.12 g/mol.

Solubility of acid in g/L = (0.0666 mol/L)*(170.12 g/mol) = 11.333 g/L (ans)